Worked solution · Exam P

SOA Exam P Sample Question #721 Solution

This Exam P sample reference tests Linear Combinations. Independence makes the aggregate variance equal to 49 × 700 squared, so the aggregate standard deviation is 4,900 while its mean is 122,500. Adding 1.281552 standard deviations gives 128.779603 thousand, which rounds to 129 and matches choice D.

Independent analysis by ActuaryProof · reviewed July 30, 2026

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Independent solution

How to solve this Linear Combinations question

Setup

Let S be the sum of all 49 independent normal amounts. A sum of independent normal variables is normal.

S=\sum_{i=1}^{49}X_i,\qquad S\sim N(\mu_S,\sigma_S^2)

\mu_S=122{,}500

Model

Add the independent component variances; the differing component means do not affect the aggregate variance.

\sigma_S^2=\sum_{i=1}^{49}\operatorname{Var}(X_i)=49(700^2)=24{,}010{,}000

\sigma_S=\sqrt{24{,}010{,}000}=4{,}900

Compute

Use the 90th-percentile standard-normal value and convert the result to thousands.

z_{0.90}=1.281551566

q_{0.90}=122{,}500+1.281551566(4{,}900)=128{,}779.6027

q_{0.90}=128.7796027\text{ thousand}

Answer

The requested percentile rounds to 129 thousand, which is choice D.

\boxed{q_{0.90}\approx 129\text{ thousand}\quad\text{(D)}}

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

ARounding the aggregate mean 122.5 thousand to 123 ignores the positive 90th-percentile shift entirely.
BA total of 125 thousand is only 0.5102 aggregate standard deviations above the mean, far below the required z_0.90 = 1.2816.
CUsing a one-standard-deviation shift gives 122.5 + 4.9 = 127.4 thousand, which rounds to 127 instead of using the 90th-percentile z-value.
EUsing the 95th-percentile value 1.6449 gives 130.56 thousand, which rounds to 131 and answers a different percentile.

Original practice · fully worked

Original variant: satellite calibration score

Three independent satellite calibration readings are A ~ N(40, 3²), B ~ N(25, 4²), and C ~ N(10, 2²). Engineers combine them into the score S = 2A − B + 0.5C. Calculate the 95th percentile of S, rounded to the nearest whole number.

A
60
B
67
C
69
D
72
E
78

Answer and calculation

The weighted score is normal with mean 60 and variance 2 squared times 9 plus 16 plus 0.5 squared times 4, which equals 53. Its 95th percentile is 60 + 1.644854 square root of 53 = 71.974715, rounding to 72 and choice D.

Setup

A linear combination of independent normal readings is itself normal.

S=2A-B+0.5C

S\sim N(\mu_S,\sigma_S^2)

Model

Apply the coefficients linearly to means and quadratically to independent variances.

\mu_S=2(40)-25+0.5(10)=60

\sigma_S^2=2^2(3^2)+(-1)^2(4^2)+0.5^2(2^2)=36+16+1=53

Compute

Use the positive 95th-percentile standard-normal value with the score's standard deviation.

z_{0.95}=1.644853627

q_{0.95}=60+1.644853627\sqrt{53}=71.97471516

Answer

The requested score percentile rounds to 72.

\boxed{q_{0.95}\approx 72\quad\text{(D)}}

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How to solve this Linear Combinations question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: satellite calibration score

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables